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Statistical Mechanics I Fall 2005 Mid-Term Quiz Review Problems

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Statistical Mechanics I Fall 2005 Mid-Term Quiz Review Problems 8.333: Statistical Mechanics I Fall 2005 Mid-term Quiz� Review Problems The Mid-term quiz will take place on Monday 10/24/05 in room 1-190 from 2:30 to 4:00 pm. There will be a recitation with quiz review on Friday 10/21/05. All topics up to (but not including) the micro-canonical ensemble will be covered. The exam is ‘closed book,’ but if you wish you may bring a two-sided sheet of formulas. The enclosed exams (and solutions) from the previous years are intended to help you review the material. Solutions to the midterm are available in the exam s section . ******** Answer all three problems, but note that the first parts of each problem are easier than its last parts. Therefore, make sure to proceed to the next problem when you get stuck. You may find the following information helpful: Physical Constants 31 27 Electron mass me 9:1 10− Kg Proton mass mp 1:7 10− Kg ∝ × 19 ∝ × 34 1 Electron Charge e 1:6 10− C Planck’s const./2α h¯ 1:1 10− J s− ∝ × 8 1 ∝ × 8 2 4 Speed of light c 3:0 10 ms− Stefan’s const. δ 5:7 10− W m− K− ∝ × 23 1 ∝ × 23 1 Boltzmann’s const. k 1:4 10− J K− Avogadro’s number N 6:0 10 mol− B ∝ × 0 ∝ × Conversion Factors 5 2 10 4 1atm 1:0 10 N m− 1A˚ 10− m 1eV 1:1 10 K ∞ × ∞ ∞ × Thermodynamics dE = T dS +dW¯ For a gas: dW¯ = P dV For a wire: dW¯ = J dx − Mathematical Formulas pλ 1 n �x n! 1 0 dx x e− = �n+1 2! = 2 R 2 2 2 � � 1 x p 2 β k dx exp ikx 2β2 = 2αδ exp 2 limN ln N ! = N ln N N −� − − − !1 − h i h i R n n ikx ( ik) n ikx ( ik) n e− = 1 − x ln e− = 1 − x n=0 n! h i n=1 n! h ic � � P 2 4 � � P 3 5 cosh(x) = 1 + x + x + sinh(x) = x + x + x + 2! 4! ··· 3! 5! ··· 2λd=2 Surface area of a unit sphere in d dimensions Sd = (d=2 1)! − 1 8.333: Statistical Mechanics I� Fall 1999 Mid-term Quiz� 1. Photon gas Carnot cycle: The aim of this problem is to obtain the blackbody ra­ diation relation, E(T; V ) V T 4, starting from the equation of state, by performing an / infinitesimal Carnot cycle on the photon gas. P P+dP T+dT P T V V+dV V (a) Express the work done, W , in the above cycle, in terms of dV and dP . Ignoring higher order terms, net work is the area of the cycle, given by W = dP dV . • (b) Express the heat absorbed, Q, in expanding the gas along an isotherm, in terms of P , dV , and an appropriate derivative of E(T; V ). Applying the first law, the heat absorbed is • @E� @E @E Q = dE + P dV = dT + dV + P dV = + P dV: @T @V @V �� �V� � �T �isotherm �� �T � (c) Using the efficiency of the Carnot cycle, relate the above expressions for W and Q to T and dT . The efficiency of the Carnot cycle (σ = dT=T ) is here calculated as • W dP dT σ = =� = : Q [(@E=@V )T + P ] T 4 (d) Observations indicate that the pressure of the photon gas is given�by P = AT , 2 4 3 where A = α kB =45 (¯hc) is a constant. Use this information to obtain E(T; V ), assuming E(0; V ) = 0. From the result of part (c) and the relation P = AT 4 , • � @E @E 4AT 4 = + AT 4 ; or = 3AT 4 ; @V� @V � �T� � �T so that E = 3AV T 4 : 2 (e) Find the relation describing the adiabatic paths in the above cycle. Adiabatic curves are given by dQ = 0, or • @E @E 0 = dT + dV + P dV = 3V dP + 4P dV; @T @V � �V � �T i.e. 4=3 P V = constant: ******** 2. Moments of momentum: Consider a gas of N classical particles of mass m in thermal equilibrium at a temperature T , in a box of volume V . (a) Write down the equilibrium one particle density feq: (p;~ ~ q ), for coordinate ~q, and mo­ mentum p~. The equilibrium Maxwell-Boltzmann distribution reads • n p2 f (p;~ ~ q ) = exp : 3=2 − 2mk T (2αmkB T ) � B � (b) Calculate the joint characteristic function, exp i~k p~ , for momentum. − · Performing the Gaussian average yields D � �E • i~k p~ mkB T 2 p˜ ~k = e− · = exp k : − 2 � � D E � � (c) Find all the joint cumulants p` pmp n . x y z c The cumulants are calculated from the characteristic function, as • � � ` m n @@ @ p`pm p n = ln p˜ ~k x y z c @ ( ikx) @ ( iky ) @ ( ikz ) � �− � �− � � − � � ��~k=0 � � � = mkB T (β`2βm0βn0 + β`0βm2βn0 + β`0βm0βn2) ; � � i.e., there are only second cumulants; all other cumulants are zero. (d) Calculate the joint moment p p (p~ p ~ ) . h � � · i Using Wick’s theorem • p p (p~ p ~ ) = p p p p h � � · i h � � π π i = p p p p + 2 p p p p h � � i h π π i h � π i h � π i 2 2 = (mkB T ) β�� βππ + 2 (mkB T ) β�π β�π 2 = 5 (mkB T ) β�� : 3 Alternatively, directly from the characteristic function, @ @@@ ~ p�p� (p~ p~ ) = p˜ k h · i @ ( ik ) @ ( ik ) @ ( ik ) @ ( ik ) � � π π �~k=0 − − − − � � � @ @ 2 mkB T ~ 2 ~ 2 � 2 k = 3mkB (mkB T ) k e�− @ ( ik ) @ ( ik ) − − � − � �~k=0 2 h i � = 5 (mkBT ) β��: � � ******** 3. Light and matter: In this problem we use kinetic theory to explore the equilibrium between atoms and radiation. (a) The atoms are assumed to be either in their ground state a0, or in an excited state a1, which has a higher energy ". By considering the atoms as a collection of N fixed two-state systems of energy E (i.e. ignoring their coordinates and momenta), calculate the ratio n1=n0 of densities of atoms in the two states as a function of temperature T . The energy and temperature of a two-state system are related by • N η E = ; 1 + exp (η/kB T ) leading to N E/η N exp (η/kB T ) E/η N 1 n0 =− = ; and n1 = = ; VV 1 + exp (η/kBT ) VV 1 + exp (η/kB T ) so that n1 η = exp : n −k T 0 �B � Consider photons θ of frequency ! = "=h¯ and momentum p~ =h!=c ¯ , which can j j interact with the atoms through the following processes: (i) Spontaneous emission: a a + θ. 1 ! 0 (ii) Adsorption: a + θ a . 0 ! 1 (iii) Stimulated emission: a + θ a + θ + θ. 1 ! 0 Assume that spontaneous emission occurs with a probability δsp, and that adsorption and stimulated emission have constant (angle-independent) differential cross-sections of δad=4α and δst=4α, respectively. (b) Write down the Boltzmann equation governing the density f of the photon gas, treating the atoms as fixed scatterers of densities n0 and n1. The Boltzmann equation for photons in the presence of fixed scatterers reads • @f @f + p~ = δadn0cf + δstn1cf + δspn1: @t · @~q − 4 (c) Find the equilibrium density feq: for the photons of the above frequency. In uniform equilibrium, the left-hand side vanishes, leaving • δ n cf + δ n cf + δ n = 0; − ad 0 eq: st 1 eq: sp 1 i.e. 1 δsp 1 δsp feq: = = : c δ n =n δ c δ exp (η/k T ) δ ad 0 1 − st ad B − st (d) According to Planck’s law, the density of photons at a temperature T depends on their 1 3 frequency ! as f = [exp (¯h!=k T ) 1]− =h . What does this imply about the above eq: B − cross sections? The result of part (c) agrees with Planck’s law if • c δ = δ ; and δ = δ ; ad st sp h3 st a conclusion first reached by Einstein, and verified later with explicit quantum mechanical calculations of cross-sections. (e) Consider a situation in which light shines along the x axis on a collection of atoms whose boundary coincides with the x = 0 plane, as illustrated in the figure. � x vacuum matter (n0 , n1 ) Clearly, f will depend on x (and px), but will be independent of y and z. Adapt the Boltzmann equation you propose in part (b) to the case of a uniform incoming flux of photons with momentum p~ =h! ¯ x=cˆ . What is the penetration length across which the incoming flux decays? In this situation, the Boltzmann equation reduces to • @f n1 px = δstc (n1 n0) f + χ (x) : @x − h3 h i To the uniform solution obtained before, one can add an exponentially decaying term for x > 0, i.e. ax=px f (px; x > 0) = A (px) e− + feq: (px) : 5 The constant A (px) can be determined by matching to solution for x < 0 at x = 0, and is related to the incoming flux. The penetration depth d is the inverse of the decay parameter, and given by px d = ; with a = δstc (n0 n1) > 0: a − ******** 6� 8.333: Statistical Mechanics I Fall 2000 Mid-term Quiz� 1. Superconducting transition: Many metals become superconductors at low temperatures T , and magnetic fields B. The heat capacities of the two phases at zero magnetic field are approximately given by 3 Cs(T ) = V �T in the superconducting phase ; 3 ( Cn (T ) = V ωT + θT in the normal phase where V is the volume, and� �, ω; θ ⎬are constants.
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